# How to determine poles and zeros of the z-transform?

This is a simple question, but I just don't understand how we determine the poles and zeros of a rational system function.

For example, for the LTI system described by this constant coefficient difference equation $$y[n]-\frac{5}{2}y[n-1]+y[n-2]=x[n]$$ we can determine that

$$H(z)=\frac{1}{(1-\frac{1}{2}z^{-1})(1-2z^{-1})}$$

I understand why there are poles at $$z=\frac{1}{2}$$ and $$z=2$$, but I don't understand why there are two zeros at $$z=0$$. Even if I multiply $$H(z)$$ through by $$z$$, there would only be one zero, correct?

## 1 Answer

Just multiply numerator and denominator by $$z^2$$ to obtain

$$H(z)=\frac{z^2}{(z-\frac12)(z-2)}$$

from which you see that there's a double zero at $$z=0$$. Note that the number of zeros and poles is always equal if you include poles and zeros at infinity.

• If I multiplied the numerator and denominator by z, then wouldn't I end up with z/((z-1/2)(z-2))? Where does the z^2 come from? – user50420 Sep 30 '18 at 17:16
• @user50420: That was a typo, I meant "by $z^2$". – Matt L. Sep 30 '18 at 17:21
• I see what you mean now. This was a trivial question, but for some reason I forgot how multiplication works. Thank for your help. – user50420 Sep 30 '18 at 17:29